Immeubles et réseaux

Le voisinage d’un sommetThe link of a vertex

Let $K$ be a field complete for a discrete valuation, with valuation ring $\mathcal O$, uniformizer $\pi$ and finite residue field $k=\mathcal O/\pi\mathcal O$; here $K=\Q_p$, $\mathcal O=\Z_p$ and $k=\F_p$. A lattice in $K^n$ is a free $\mathcal O$-submodule of rank $n$, and two lattices are homothetic when one is a multiple of the other. The building of $\PGL(n,K)$ has the homothety classes $[L]$ as vertices, two of them adjacent when they have representatives with $\pi L\subsetneq L'\subsetneq L$, and the sets of pairwise adjacent vertices as simplices. For $n=2$ it is a tree in which every vertex has $|k|+1$ neighbours.

The neighbours of a vertex are read off modulo $\pi$, so a vertex sees a finite projective space: the eight points of $\Proj^1(\F_7)$ in the tree over $\Q_7$, the incidence graph of the Fano plane in the building over $\Q_2$. All three parts of the theorem were also confirmed by computation with rational basis matrices, for $(n,p)=(3,2)$, where the link is the Heawood graph, with 14 vertices, cubic, of girth 6 and diameter 3; for $(3,3)$, with 26 vertices of degree 4; and for $(2,7)$, eight neighbours, no two adjacent, with the Möbius action even for $\mathrm{diag}(3,1)$, whose determinant is not a square.

A completion of a finite projective geometry $\Pi$ over $k$ is a building of type $\tilde A_{n-1}$ with a vertex whose link is the flag complex of $\Pi$. The building lives over $K$, a completion of a global field, and the finite geometry over its residue field. Completions are not unique: $\Q_2$ and $\F_2(\!(t)\!)$ both have residue field $\F_2$, so each gives a completion of the Fano plane, and the chapter finds them glued in different ways.

Une double vie en deux sommetsA double life at two vertices

Read with $(n,p)=(3,2)$ and with $(n,p)=(2,7)$, the theorem on vertex links puts the two lives of the group of order 168 at two vertices. In the building of $\PGL(3,\Q_2)$ a vertex sees the Fano plane and the group is the residue of $\PGL(3)$; in the tree of $\PGL(2,\Q_7)$ a vertex sees $\Proj^1(\F_7)$ and the group is the residue of $\PGL(2)$.

The link at 2 has more symmetry than the group. Its automorphisms number 336, and the 168 that do not preserve the two types of vertex are the dualities, exchanging points and lines. On the line side they become the Möbius maps of non-square determinant: the isomorphism of the full automorphism groups, $\Aut(\text{Heawood})\cong\PGL(2,7)$, extends the double life by the dualities.

Une complétion recollée par les octonionsA completion glued by the octonions

A building of type $\tilde A_2$ with a group acting simply transitively on its vertices is described by finite data, a triangle presentation in the sense of Cartwright, Mantero, Steger and Zappa. Let $(P,\mathcal L)$ be a finite projective plane and $\lambda\colon P\to\mathcal L$ a bijection. A triangle presentation compatible with $\lambda$ is a set $T$ of triples of points such that (A1) some $(x,y,z)$ lies in $T$ exactly when $y\in\lambda(x)$, (A2) $(x,y,z)\in T$ implies $(y,z,x)\in T$, and (A3) for $x,y$ there is at most one $z$ with $(x,y,z)\in T$. Its group is $\Gamma_T=\langle a_x\mid a_xa_ya_z=1\text{ for }(x,y,z)\in T\rangle$. By their theorem the Cayley graph of $\Gamma_T$ is the 1-skeleton of a thick building of type $\tilde A_2$, on whose vertices $\Gamma_T$ acts simply transitively, with every link the incidence graph of $(P,\mathcal L)$.

The octonion table is such data. Take the imaginary units $e_x$, $x\in\Z/7$, with $e_xe_{x+1}=e_{x+3}$, its rotations, $e_x^2=-1$ and $e_be_a=-e_ae_b$. Its 21 oriented triples $T_{\Oct}=\{(a,b,c):e_ae_b=+e_c\}$ are the rotations of $(x,x+1,x+3)$; the sets $\lambda(x)=x+\{1,2,4\}$ are the lines of a Fano plane, since $\{1,2,4\}$ is a perfect difference set modulo 7; and $T_{\Oct}$ satisfies (A1)–(A3). The permutations of $\Z/7$ preserving it are the 21 maps $x\mapsto ax+b$ with $a\in\{1,2,4\}$, and the abelianization of $\Gamma_{\Oct}$ is $\Z/2\times\Z/2\times\Z/6$, from the Smith normal form of the circulant relation matrix $I+S+S^3$, of determinant 24. So $\Gamma_{\Oct}$ acts simply transitively on a building in which the triangles at each vertex are the oriented quaternionic triples of the octonions.

The bridge from the octonion table to this triangle presentation is built, since the two share their data, the set of oriented triples. Nothing more is identified. The octonions are not a group, and $a_x\mapsto e_x$ is not a homomorphism: $e_xe_{x+1}e_{x+3}=-1$, while the group imposes $a_xa_{x+1}a_{x+3}=1$. In the classification of the triangle presentations of order 2 the table is A.1, Example 2 of Vaes and Valvekens, whose building is quoted as that of $\F_2(\!(t)\!)$; the theorem confirms this independently.

Le recollement de MumfordMumford’s gluing

Mumford built his fake projective plane from a lattice $\Gamma_M$ in $\PGL(3,\Q_2)$, discrete, cocompact and transitive on the vertices of the building. Following Kato, let $E=\Q(\sqrt{-7})$ and $\alpha=(-1+\sqrt{-7})/2$, so that $\mathcal O_E=\Z[\alpha]$ and $2=\alpha\bar\alpha$, and let $L=\Q(\zeta)$, $\zeta=e^{2\pi i/7}$, so that $\alpha=\zeta+\zeta^2+\zeta^4$. On $L$, an $E$-space with basis $1,\zeta,\zeta^2$, the form $h(x,y)=\operatorname{tr}_{L/E}(x\bar y)$ is positive definite of determinant 7; modulo $\sqrt{-7}$ it has rank one, with null plane $N=\{x_1+x_2+x_3=0\}$ of $\F_7^3$. Let $\Gamma_1$ be the image in $\PGL(3,\Q_2)$ of the similitudes of $h$ that preserve $\mathcal O_L\otimes\Z_\ell$ for every prime $\ell\neq2$, and $\Gamma_M$ the image of those whose action on $N$ lies in a fixed Sylow 2-subgroup of the elements of $\GL(2,\F_7)$ of determinant $\pm1$.

The stabilizer of $v_0=[\Z_2^3]$ in $\Gamma_1$ is the Frobenius group of order 21 generated by multiplication by $\zeta$ and by $\zeta\mapsto\zeta^2$; reduction maps $\Gamma_1$ onto $\PSL(2,7)$, and $\Gamma_M$ is the preimage of a Sylow 2-subgroup, dihedral of order 8, with trivial stabilizer at $v_0$. For each of the seven neighbours $[L_V]$, $V$ a plane of $\F_2^3$, exactly one $a_V\in\Gamma_M$ moves $v_0$ there, and the triples with $a_Ua_Va_W=1$ form a triangle presentation $T_M$: with a suitable numbering, the rotations of $(0,0,1)$, $(0,2,3)$, $(1,3,4)$, $(1,5,2)$, $(2,4,6)$, $(3,6,5)$, $(4,5,6)$, with the lines $\{0,1,2\}$, $\{0,3,5\}$, $\{1,3,4\}$, $\{0,4,6\}$, $\{1,5,6\}$, $\{2,3,6\}$, $\{2,4,5\}$ for $x=0,\dots,6$. Its group is $\Gamma_M$, and up to relabelling $T_M$ depends neither on the Sylow subgroup nor on the conventions, the prime $(\alpha)$ or $(\bar\alpha)$ and $H$ or its transpose. All of this was computed in exact arithmetic in $\mathcal O_E$.

Two finite invariants separate the gluings: the abelianization of $\Gamma_M$ is $\Z/2\times\Z/6$, and only the identity relabelling preserves $T_M$, against 21 for the octonion table. Nor is $T_M$ Example 1 of Vaes and Valvekens, whose building is also that of $\Q_2$ and whose group has abelianization $\Z/3$. A stronger statement rests on the fields, not on a computation: no subgroup of finite index in $\Gamma_{\Oct}$ is isomorphic to one in $\Gamma_M$. Isomorphic subgroups would make the two buildings quasi-isometric (Švarc–Milnor), a quasi-isometry would induce an isometry of their Tits boundaries (Kleiner–Leeb), the incidence graphs of the projective planes over $\F_2(\!(t)\!)$ and $\Q_2$, and a collineation or correlation between them would make the fields isomorphic, against their characteristics 2 and 0. The octonion table glues with the symmetry $x\mapsto2x$; Mumford’s gluing has none, and the symmetry of order 21 sits instead in the vertex stabilizer of $\Gamma_1$, which meets $\Gamma_M$ trivially. In $\rho(\Gamma_{\Oct})\rtimes F$, with $F$ generated by $v\mapsto wv$ and $v\mapsto v^2$, which conjugate $M_x$ to $M_{x-1}$ and to $M_{2x}$, the lattice is a normal subgroup; in $\Gamma_1$, where $\Gamma_M$ is its own normalizer with 21 conjugates, no normal subgroup acts simply transitively. At a vertex the two actions look alike, both stabilizers acting on the Fano link as the normalizer of a Sylow 7-subgroup of $\GL(3,2)$. Globally one stabilizer has a normal complement and the other has none, and the theorem says why.

La place archimédienneThe archimedean place

Let $\zeta=(1+\sqrt{-3})/2$, so that $\zeta^2=\zeta-1$, and $\mathcal O=\Z[\zeta]$, the Eisenstein integers, a Euclidean domain. $\PSL(2,\mathcal O)$ is a discrete subgroup of finite covolume of $\PSL(2,\C)$, acting on hyperbolic space $\mathbb H^3$ with sphere at infinity $\Proj^1(\C)$, and its cusps are the points of $\Proj^1(\Q(\sqrt{-3}))$. The element $2+\zeta$ has norm 7; let $\mathfrak p=(2+\zeta)$ and $\Gamma(\mathfrak p)$ the kernel of reduction $\PSL(2,\mathcal O)\to\PSL(2,\mathcal O/\mathfrak p)$.

Since $7=(2+\zeta)(2+\bar\zeta)$ splits, the completion at $\mathfrak p$ is $\Q_7$, and $\Proj^1(\F_7)$ appears twice for the one field: at the finite place as the link of a vertex of the tree, at the complex place as the cusps of the congruence quotient. One group sees both. $\Gamma_S=\PSL(2,\mathcal O[1/\pi])$, $\pi=2+\zeta$, acts on $\mathbb H^3$ and on the tree of $\PGL(2,\Q_7)$; the stabilizer of the base vertex is $\PSL(2,\mathcal O)$, the elements fixing its eight neighbours form $\Gamma(\mathfrak p)$, and $[v]\mapsto[v\bmod\mathfrak p]$ identifies the cusps with the neighbours, a primitive $v$ giving the half-line of lattices $\mathcal O_{\mathfrak p}v+\mathfrak p^nL_0$.

It is also a congruence quotient of two arithmetic groups at primes over 7: of $\PSL(2,\mathcal O)$, a lattice in $\PSL(2,\C)$, and of Mumford’s $\Gamma_1$, a lattice in $\PGL(3,\Q_2)$ made of rational points of a group whose real form is compact. Both reductions give the line life, $\PSL(2,7)$ on the eight cusps and on the eight lines of $N$, and in $\Gamma_1$ the plane life is present too, as the link at 2. This is not a double life, since both parents reduce to the same life. It is one finite life with two arithmetic parents of opposite real type.

Les cellules du complément d’entrelacs de congruenceThe cells of the congruence link complement

The tessellation of $\mathbb H^3$ by regular ideal tetrahedra with a face on $\infty,0,1$ has orientation-preserving symmetry group $\PGL(2,\mathcal O)$, in which $\Gamma(\mathfrak p)$ is normal. So it descends to $M$, and $\PGL(2,7)$ acts on $M$ preserving it, with $\PSL(2,7)$ the deck group. Identify the cusps with $\Proj^1(\F_7)$ so that $\zeta$ goes to 5; the tetrahedron $\tau_0=\{\infty,0,1,\zeta\}$ becomes $\{\infty,0,1,5\}$. Every cell is determined by its cusps, so every configuration of cells is a configuration of points of the line, and its stabilizer can be computed there. Every row of the seam table then has an incarnation in the cells of one manifold: a cusp with a face through it (1), an edge with a tetrahedron through it ($C_2$, one orbit for each class), a face ($C_3$), four cusps that are not the cusps of a tetrahedron ($C_4$), a tetrahedron of class $a$ or $b$ with a pair of opposite edges ($V_4^a$, $V_4^b$), an edge ($S_3$), a cusp with a class of parallel edges of its cusp torus ($C_7$), a partition of the cusps into two fours that are not tetrahedra ($D_8$), a tetrahedron of class $a$ or $b$ ($A_4^a$, $A_4^b$), a cusp ($7{:}3$), a complementary pair of tetrahedra of class $a$ or $b$ ($S_4^a$, $S_4^b$), and $M$ itself ($G$).

The two sets of 28 in $M$ are different objects. The edges are the object of size 28; the tetrahedra are two objects of size 14, whose stabilizers $A_4$ are not even isomorphic to the edges’ stabilizers in $\PGL(2,7)$, dihedral of order 12. So the type bridge between the 28 tetrahedra and the object of size 28, with equal size and one group, is refuted; through the Fano column of the table the tetrahedra are the complete quadrangles and quadrilaterals, each with one of its two orientations. Sending an edge to its two cusps is the unique seam from the edges to the 2-subsets, and composing gives the Sylow 3-subgroup fixing them, the antiflag it fixes, the bitangent of the Klein quartic through the two points it fixes, and a vertex of the Coxeter graph; two edges are adjacent in the Coxeter graph exactly when their pairs are disjoint and harmonic. For the edge from 0 to $\infty$ the subgroup is generated by $z\mapsto2z$ and the bitangent is $x+y+z=0$.

A cross-section of the cusp at $\infty$ is $\C/\mathfrak p$, triangulated by the 7 edges and 14 tetrahedra there: a torus with 7 vertices, 21 edges and 14 triangles whose 1-skeleton is $K_7$. Labelling the end of the edge from $\infty$ to $a$ by $a$, the triangles are $a+\{0,1,5\}$ and $a+\{0,4,5\}$. The second family is the family of lines $x+\{1,2,4\}$ of the octonion presentation, the first its mirror image, and the stabilizer of the cusp acts on the labels as the maps $x\mapsto ax+b$, $a\in\{1,2,4\}$. The bridge to the octonion completion is built for this shared configuration, through $\zeta\mapsto5$, and for nothing more.